Find the `13^(t h)`term in the expansion of `(9x-1/(3sqrt(x)))^(18),x!=0`

Explanation

To find the 13th term in the expansion of \((9x - \frac{1}{3\sqrt{x}})^{18}\), we will use the Binomial Theorem. The general term in the expansion of \((p + q)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

### Step-by-step Solution:

1. **Identify \(p\), \(q\), and \(n\)**:
- Here, \(p = 9x\), \(q = -\frac{1}{3\sqrt{x}}\), and \(n = 18\).

2. **General Term**:
- The \(r^{th}\) term (or \(T_{r+1}\)) in the expansion can be expressed as:
\[ T_{r+1} = \binom{18}{r} (9x)^{18-r} \left(-\frac{1}{3\sqrt{x}}\right)^r \]

3. **Finding the 13th Term**:
- The 13th term corresponds to \(r = 12\) (since \(T_{r+1}\) means \(r\) starts from 0).
\[ T_{13} = \binom{18}{12} (9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}\right)^{12} \]

4. **Simplifying the Expression**:
- Calculate \(\binom{18}{12}\):
\[ \binom{18}{12} = \binom{18}{6} = \frac{18!}{12! \cdot 6!} = 18564 \]
- Calculate \((9x)^6\):
\[ (9x)^6 = 9^6 \cdot x^6 \]
- Calculate \(\left(-\frac{1}{3\sqrt{x}}\right)^{12}\):
\[ \left(-\frac{1}{3\sqrt{x}}\right)^{12} = \frac{(-1)^{12}}{3^{12} \cdot (\sqrt{x})^{12}} = \frac{1}{3^{12} \cdot x^6} \]

5. **Combine the Terms**:
- Now substitute back into the expression for \(T_{13}\):
\[ T_{13} = 18564 \cdot 9^6 \cdot x^6 \cdot \frac{1}{3^{12} \cdot x^6} \]
- The \(x^6\) terms cancel out:
\[ T_{13} = 18564 \cdot \frac{9^6}{3^{12}} \]

6. **Simplifying \(9^6\) and \(3^{12}\)**:
- Note that \(9 = 3^2\), so:
\[ 9^6 = (3^2)^6 = 3^{12} \]
- Therefore:
\[ T_{13} = 18564 \cdot \frac{3^{12}}{3^{12}} = 18564 \]

### Final Answer:
The 13th term in the expansion is \(18564\).